Question

A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10^–5 K^–1; Young’s modulus of brass = 0.91 × 10¹¹ Pa.

Answer 1

nitial temperature, T₁ = 27°C

Length of the brass wire at T₁, l = 1.8 m

Final temperature, T₂ = –39°C

Diameter of the wire, d = 2.0 mm = 2 × 10^–3 m

Tension developed in the wire = F

Coefficient of linear expansion of brass, α= 2.0 × 10^–5 K^–1

Young’s modulus of brass, Y = 0.91 × 10¹¹ Pa

Young’s modulus is given by the relation:

${\displaystyle Y=\frac{\text{Stress}}{\text{Strain}}=\frac{\frac{F}{A}}{\frac{△L}{L}}}$

${\displaystyle △L=\frac{F\times L}{A\times Y}}$   ...(i)

Where,

F = Tension developed in the wire

A = Area of cross-section of the wire.

ΔL = Change in the length, given by the relation:

ΔL = αL(T₂ – T₁) … (ii)

Equating equations (i) and (ii), we get:

${\displaystyle \alpha L(T_2-T_2)=\frac{FL}{\pi {\left(\frac{d}{2}\right)}^2\times y}}$

${\displaystyle F=\alpha (T_2-T_1)\pi Y{\left(\frac{d}{2}\right)}^2}$

${\displaystyle F=2\times {10}^{-5}\times (-39-27)\times 3.14\times 0.91\times {10}^{11}\times {\left(\frac{2\times {10}^{-3}}{2}\right)}^2}$

${\displaystyle =-3.8\times {10}^{2}N}$

(The negative sign indicates that the tension is directed inward.)

Hence, the tension developed in the wire is 3.8 ×10² N.

Answer 2

Here l = 1.8 m

${\displaystyle △t =(-39-27){}^{\circ }C=-66{}^{\circ }C}$

${\displaystyle \alpha =2.0\times {10}^{-5}{K}^{-1}}$

${\displaystyle Y =0.91\times {10}^{11}Pa}$

${\displaystyle A=\frac{\pi D^2}{4}=\frac{22}{7}\times \frac{1}{4}{(2\times {10}^{-3})}^2{m}^2}$

Now, ${\displaystyle Y=\frac{\text{Fl}}{A△L}\Rightarrow △l= \frac{Fl}{AY}\text{or}l\alpha △t= \frac{Fl}{AY}}$

or ${\displaystyle F=-YA\alpha △t}$

or ${\displaystyle F=-0.91\times {10}^{11}\times \frac{22}{7}\times \frac{1}{4}{(2\times {10}^{-3})}^2\times 2.0\times {10}^{-5}\times 66N}$

${\displaystyle =-3.77\times {10}^{2}N}$